1. Module B3
3.1 Sinusoidal steady-state analysis (single-phase), a review
3.2 Three-phase analysis
Kirtley
Chapter 2: AC Voltage, Current and Power
2.1 Sources and Power
2.2 Resistors, Inductors, and Capacitors
Chapter 4: Polyphase systems
4.1 Three-phase systems
4.2 Line-Line Voltages 1

2. Basic concepts of electric power
With exception of a few High Voltage DC circuits, the entire electric grid operates to deliver “alternating current” (AC). An “AC voltage” appears below.
The maximum value of the waveform is 400 volts.
The waveform completes one cycle in sec. This is the “period” of the waveform and is 1/f. So, f=1/0.0167=60Hz or 60 cycles/sec.
Since 1 cycle completes 2π radians, ω=(2π rad/cyc)*60cyc/sec=377rad/sec.
We can therefore express this voltage as 2

3. KVL and Ohm’s Law
The below is a circuit model of a voltage source connected to a resistance.
We apply Kirchoff’s Voltage Law (KVL): the sum of voltages around the loop must be 0:
We may apply Ohm’s Law to obtain the current: 3

4. Voltage and current for resistive cct
Compare voltage & current. Assume R=2 ohms.
Current & voltage cross 0 simultaneously 4

5. Capacitance
The below is a circuit model of a voltage source connected to a capacitance.
The current is given by:
Again, from KVL: 5

6. Voltage and current for capacitive cct
Compare voltage & current. Assume C=0.001farad.
Current crosses 0 before voltage. We say “current leads voltage.”
If we multiply the time axis by ω=377 rad/sec, we would see that current leads voltage by π/2 radians. 6

7. Inductance
The below is a circuit model of a voltage source connected to an inductor. 3 3
Again, from KVL:
The current is given by: 7

8. Voltage and current for inductive cct
Compare voltage & current. Assume L=.01 henry.
Current crosses 0 after voltage. We say “current lags voltage.”
If we multiply the time axis by ω=377 rad/sec, we would see that current lags voltage by π/2 radians. 8

9. Impedance It is clear that ZR=R. Is ZC=1/ωC? Is ZL=ωL?
In the three expressions below, we use Ohm’s law to express current in terms of impedance, then equate that expression to the one we found in the 5 previous slides.
It is clear that ZR=R.
RESISTIVE:
CAPACITIVE:
INDUCTIVE:
Is ZC=1/ωC?
Is ZL=ωL?
No, because we also need to take care of the phase shifts. 9

10. Impedance
To do so, let’s utilize the imaginary unit “j” which has the following properties:
It is the square root of -1:
From the Euler relation:
With ϕ=π/2, we obtain:
Therefore, j has magnitude of “1” and angle π/2.
Therefore
Then, with ZC=1/jωC, ZL=jωL, we get correct current expressions. 10

11. Impedance
RESISTIVE:
CAPACITIVE:
INDUCTIVE:
Z, called impedance, is a generalization of resistance that accounts for the effects of capacitance & inductance. 11

12. Impedance In a purely resistive circuit, Z=ZR=R.
In a purely capacitive circuit, Z=ZC=1/jωC.
In a purely inductive circuit, Z=ZL=jωL.
We also define:
Capacitive reactance: XC=1/ωC
Inductive reactance: XL=ωL
In a purely resistive circuit, Z=ZR=R.
In a purely capacitive circuit, Z=ZC=1/jωC=XC/j.
But recall j=√-1, then j2=-1. So (XC/j)*(j/j)=-jXC.
So Z=ZC=-jXC.
In a purely inductive circuit, Z=ZL=jωL=jXL. 12

13. Impedance
It is possible to have an impedance that is a combination of resistance, capacitance, and inductance. 3
Then
The impedance, Z, can be thought of as a vector on the complex plane. 13

14. Impedance
Imaginary
Inductive
Real
Capacitive 14

15. Example 1 Let R=2Ω, C=0.001farad, L=0.01henry, Vmax=400. 3
Then Z=R+jωL-j/ωC
The current will be
But how to get this? 15

16. Example 1
You can divide each numerical point in the sinusoidal numerator by the complex denominator, then take the magnitude and angle of the resulting complex number.
But observe that only the magnitude changes as t increases. This is because the “phasor angle” of the numerator and the vector angle of the denominator both remain constant with t. That is:
But what is |Z|∟θZ? 16

17. Impedance
Imaginary
Real
So:
And from the previous slide: 17

18. Impedance So
Recalling that 28.92°= radians, we write the above as:
We can plot this:
Current crosses 0 after voltage. We say “current lags voltage.”
If we multiply the time axis by ω=377 rad/sec, we would see that current lags voltage by radians. 18

19. Phasors Let’s review what we just did:
Ohm’s Law
Express vL(t)
Express numerically
Express in “polar” notation
Separate magnitude & angle calculations
Express numerically
Do the arithmetic
Convert from polar notation to trigonometric notation
Notice that nothing ever happened to the ωt part. Nor did the sinusoidal function change. This means that the frequency of the current is the same as the frequency of the voltage. If this is the case, then why do we need to carry the sinωt through the calculations? Why not just drop it? This is what we do when we use phasor notation, as in the next slide. 19

20. Phasors Define: Phasor for source voltage: Phasor for load voltage:
Phasor for current:
Also recall the impedance:
Then…
There is an implicit sinωt that we are not writing. This sinωt gives rotation to the vectors which make them phasors.
There is no implicit sinωt  impedance is a vector, not a phasor!
Rotation is in CCW direction
Imaginary
These do rotate.
Real Z
(does not rotate) 20

21. Instantaneous Power Instantaneous power is given by Let Then
Apply trig identity:
Gathering terms in the argument of the second cos function: 21

22. Instantaneous Power Apply trig identity to last term:
Rearrange and distribute:
Then:
where:
Real power
Active power
Watts
Imaginary power
Reactive power
VARs 22

23. Example 2 Consider example 1, where Vmax=400, θv=0,
and we computed that Imax=175.06, θi = rad,
Then:
watts
vars
TERM1
TERM2
TERM3 23

24. Example 2
TERM1
TERM2
TERM3
The average of a periodic signal x(t) is given by:
Power is delivered only if the average power is not 0.
The average value of term2 and term3 is 0. These terms are not responsible for power delivery.
Average value of the composite [the composite is instntans power, p(t)] is term1, P.
Interesting side note: instntns power varies at twice the nominal frequency. 24

25. Effective or RMS value
We want to obtain the constant (or DC) value of a current which delivers the same average power to a resistor.
AC circuit
DC circuit
The power delivered is:
The power delivered is:
Equate 25

26. Effective or RMS value
We want to obtain the constant (or DC) value of a voltage which delivers the same average power to a resistor.
AC circuit
DC circuit
The power delivered is:
The power delivered is:
Equate 26

27. Effective or RMS value
The root-mean-square value of any periodic signal x(t) is:
Let x(t) be v(t):
Apply trig identity:
Current may be similarly treated, resulting in 27

28. Power and Phasor notation
We previously derived power expressions for P and Q in terms of Vmax and Imax. We may now express these in terms of rms (effective) values:
This simplifies our power expressions. Commercial voltages and currents are always specified as rms values.
Phasor magnitudes may be rms or maximum. Unless it is otherwise specified, we assume phasor magnitudes are rms, and we will drop the subscript “eff.” Thus, 28

29. Power factor
Power factor angle, defined θ, is the angle by which voltage across an element leads current through the element, i.e., θ=θv-θi. It is also the angle of the load Z, i.e., Z=|Z|∟θ. One observes from the P, Q expressions that it is also the argument of the trig functions. Therefore:
The power factor is given as pf=cosθ. It is always positive, therefore we usually indicate pf with either the word “leading” (for capacitive load) or “lagging” (for inductive load). The following table summarizes important relations.
Load, Z
Im{Z}
θi (θv=0)
θ (θv=0) pf
Q sign
Q Load
R only
1.0
0 vars
R and L + -
0 to 1, lagging
Absorbing vars
R and C
0 to 1, leading
Supplying vars 29

30. Power factor
Consider a load at the end of a transmission line. Assume the load is highly inductive. This means current is lagging, θi<0, and θ>0 (with θv=0). 3
Trans Line impedance
Load
Assume V is constant, and we must supply P (this is the real power of the load). With V and P constant, from P=VIcosθ, Icosθ=I×pf must be constant. What do we want pf to be? 30

31. Power factor 3 Assume P=33MW, V=39.84kV. This means that
Here are a few possibilities for Icosθ =I×pf.
I (amperes)
828.3
871.9
920.3
1035.4
Pf (leading) 1
0.95
0.9
0.8
We need higher power factor to obtain lower current.
Lower current is desired in order to minimize
I2R (real power) losses in the transmission line. 3
Trans Line impedance
Load 31

32. Power factor correction (PFC)
So what do you do if your load has a power factor of 0.85 but your provider requires that you have 0.95 ?
Install a PFC device at the load’s point of interconnection with the grid. 3
Trans Line impedance
Power factor correction device
Load
This least expensive PFC device is a switched capacitor, or several switched capacitors. These are static devices, however. Dynamic devices are also available but are more expensive (and you get dynamic responsiveness as well). 32

33. What is reactive power? 1. It is given by:
2. It occurs as the coefficient of the double frequency “sin” term in p(t):
3. It is the maximum value of the rate of energy flowing alternately into the reactive component of the load (away from the source) and away from the reactive component of the load (toward the source). It is only present when there is a stored energy device (C or L) in the circuit.
4. Its units are “volt-amperes-reactive” or “vars.” Dimensionally, vars are the same as watts and volt-amperes.
5. Three attributes of reactive power:
It does no useful work (its average value is 0).
Its presence increases current magnitude for a given P and V.
Derivatives of Q “flow” equations (not shown in these slides) indicate that Q is closely related to voltage magnitudes. 33

34. Complex power Complex power is defined as: where:
The asterisk next to the “I” phasor indicates conjugation: 34

35. Power triangle
|S| Q θ P 35

36. Example 3  θi=-θ=30°. Thus, P=0.482100MW, Q=-0.27834MVARs
The below represents one of the phases of a generator supplying a load. The box to the right represents the load. The current flowing into the load is A at pf leading, with volts at its terminals. What complex power is delivered to the load?
Solution: θ=-cos-1(.866)=-30°.
 θi=-θ=30°.
Thus, P= MW,
Q= MVARs
We also define the apparent power |S|, the magnitude of S: 36